The minimum distance between an object and its real image formed by a convex lens of focal length $f$ is

We know: Let the seperation between object and image by $x$, $$ x=u-v $$ Using equation (1), on rewriting, $$ \begin{aligned} & x=u-\frac{x f}{u} \\ & \Rightarrow x=\frac{u^2}{(u+f)}---(1) \end{aligned} $$ we first take log of equation (1) $$ \Rightarrow \ln (x)=2 \ln (u)-\ln (u+f) $$ Subsequently, to minimize $x$ w.r.t. $u$, take the derivative of $x$ w.r.t. $u$ : $$ \begin{aligned} & \Rightarrow \frac{1}{x} \frac{d x}{d u}=\frac{2}{u}-\frac{1}{(u+f)} \\ & \Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}---(2) \end{aligned} $$ Condition for minimum $x$ requires, $\frac{d x}{d u}=0$ and $\frac{d^2 x}{\partial u^2}<0$ $$ \Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}=0 $$ So, $x=0$ \{is the trivial solution $\}$ and the real solution occurs for $u=-2 f=-R$. Lets check the minimum value using equation (1), $$ \Rightarrow x_{\min }=\frac{u^2}{(u+f)}=\frac{(-2 f)^2}{(-2 f+f)}=4 f $$